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AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 6: Energy & Momentum of Rotating SystemsAngular Momentum & Angular ImpulseAngular Impulse Graphs

Angular Impulse Graphs (College Board AP® Physics 1: Algebra-Based) : Study Guide

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

Torque-time graph

  • A changing torque can be plotted as a function of time

  • The angular impulse delivered to an object or rigid system is equal to the product of torque and time

  • Therefore, the area under the torque-time graph is equal to

    • angular impulse

    • change in angular momentum

area space equals space tau increment t space equals space angular space impulse space equals space increment L

Determining angular impulse from a torque-time graph

Graph of torque in newton meters (Nm) against time in seconds (s) with a label indicating that the area under the curve represents the angular impulse or change in angular momentum
When the torque is not constant, the angular impulse is the area under a torque–time graph

Worked Example

The graph shown represents the torque exerted on an object as a function of time over a 5-second interval.

Graph of torque (τ) vs. time (t) showing values in N m. Torque rises from 0 to 10 between 0 and 2 seconds, drops to 0 by 3 seconds, then stays at -5 until 5 seconds.

The object has a rotational inertia of 5.0 kg·m2. At t = 0 s, the object has an angular velocity of −2.0 rad/s.

Determine the angular velocity of the object at t = 5 s.

Answer:

Step 1: Analyze the scenario

  • From t = 0 to t = 2 s, the torque increases to a maximum of 10 N·m, and then, from t = 2 s to t = 3 s, the torque decreases back to 0 N·m

  • At t = 3 s, the direction of the applied torque changes

  • From t = 3 to t = 5 s, the torque remains at a value of −5 N·m

  • Overall, the angular impulse delivered to the object:

    • between t = 0 to t = 3 s is positive, so the angular velocity will increase

    • between t = 3 to t = 5 s is negative, so the angular velocity will decrease

Step 2: Determine the change in angular momentum over the 5 s interval

  • The area under a torque-time graph is equal to angular impulse, or the change in angular momentum

Graph depicting torque (τ) in N·m versus time (t) in seconds, showing positive and negative angular impulses. The positive impulse (0-3s) is a triangle, the negative (3-5s) is a rectangle.

  • Area under the positive curve (triangle) = 1 half cross times 10 cross times 3 space equals space 15 space straight N times straight m times straight s

  • Area under the negative curve (rectangle) = negative 5 cross times 2 space equals space minus 10 space straight N times straight m times straight s

  • Therefore, the angular impulse, or change in angular momentum is:

increment L space equals space 15 space minus space 10 space equals space 5 space straight N times straight m times straight s

Step 3: Determine the final angular velocity of the object

  • The change in angular momentum is equal to

increment L space equals space I increment omega space equals space I open parentheses omega space minus space omega subscript 0 close parentheses

  • Where I = 5.0 kg·m2 and omega subscript 0 = −2.0 rad/s.

  • Therefore, when t = 5 s, the angular velocity is:

5 space equals space 5.0 open parentheses omega space minus space open parentheses negative 2 close parentheses close parentheses

omega space equals space minus 1.0 space rad divided by straight s

Angular momentum-time graph

  • A changing angular momentum can be plotted as a function of time

  • The net torque exerted on a system is equal to the rate of change of angular momentum

  • Therefore, torque is equal to the slope of an angular momentum-time graph

slope space equals space fraction numerator L subscript 2 space minus space L subscript 1 over denominator t subscript 2 space minus space t subscript 1 end fraction space equals space fraction numerator increment L over denominator increment t end fraction space equals space tau subscript n e t end subscript

Determining torque from an angular momentum-time graph

Graph of angular momentum in kg·m2/s against time in s, with a label indicating that the slope represents torque
When angular momentum varies with time, the net torque exerted on the system can be found from the slope of the angular momentum-time graph

Worked Example

The graph shown represents the angular momentum of the two objects, A and B, as functions of time between time t = 0 s and t = 5 s.

The average magnitudes of the net torques on objects A and B from t = 0 s to t = 5 s are tau subscript A and tau subscript B, respectively.

Graph showing the angular momentum (L) versus time (t) for Object A (solid line) and Object B (dashed line). Object A: between t = 0 to t = 2 s, angular momentum increases from 0 to 10 kgm2/s, between t = 2 to t = 3 s, angular momentum does not increase, between t = 3 to t = 5 s, angular momentum increases from 10 to 20 kgm2/s. Object B: between t = 0 to t = 3 s, angular momentum increases from 0 to 5 kgm2/s, between t = 3 to t = 5 s, angular momentum does not increase.

Which of the following expressions correctly relates the magnitudes of the average torques?

A      tau subscript B space equals space 4 tau subscript A

B      tau subscript B space equals space 2 tau subscript A

C      tau subscript B space equals space 1 half tau subscript A

D      tau subscript B space equals space 1 fourth tau subscript A

The correct answer is D

Answer:

Step 1: Analyze the scenario

  • Object A and B experience a varying net external torque over the 5 second interval

  • The average magnitude of the net torque on each object is given by:

tau subscript a v g end subscript space equals space fraction numerator increment L over denominator increment t end fraction

  • Where increment L is the change in angular momentum from t = 0 s to t = 5 s

Step 2: Determine the change in angular momentum for each object

  • Both A and B have L = 0 kg·m2/s at t = 0 s

  • For object A:

    • angular momentum: L subscript A = 20 kg·m2/s at t = 5 s

    • change in angular momentum: increment L subscript A = 20 kg·m2/s

  • For object B:

    • angular momentum: L subscript B = 5 kg·m2/s at t = 5 s

    • change in angular momentum: straight capital delta L subscript B = 5 kg·m2/s

Step 3: Determine the average torque on each object

  • The average torque on object A is:

tau subscript A space equals space 20 over 5 space equals space 4 space straight N times straight m

  • The average torque on object B is

tau subscript B space equals space 5 over 5 space equals space 1 space straight N times straight m

Step 4: Determine the ratio of the average torques

  • The ratio of the average torque on object B to object A is

tau subscript B over tau subscript A space equals space 1 fourth

tau subscript B space equals space 1 fourth tau subscript A

  • Therefore, the correct option is D

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    Author
    Reviewer
    Katie M

    Author: Katie M

    Expertise: Physics Content Creator

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    Caroline Carroll

    Reviewer: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.